# CDF of Piecewise Folded Normal

I came across a problem in a Carmona's Statistical Analysis of Financial Data in R (pg. 189, Problem 3.13). The due date has passed, so now it is considered a self-study question.

I am seeking a simple-to-understand proof to the following:

$$X_1 \sim \mathcal{N}(0,1)$$, $$X_2 \sim \mathcal{N}(0,1)$$

$$X_3 \sim \begin{cases} |X_2| & X_1 \geq 0 \\ -|X_2| & X_1 < 0 \end{cases}$$

(Q.) Compute the CDF of $$X_3$$. State if $$X_3$$ is Gaussian.

Intuitively, I believe $$X_3$$ should be standard normal $$\sim \mathcal{N}(0,1)$$. Half of the time, $$X_3 =$$ folded normal, while half of the time $$X_3 =$$ negative of a folded normal, by simple symmetry. I run an R simulation to test my suspicions and find this to be correct (the R code and plot is listed at the end of the question).

We know $$\mathbb{P}(X_1 \leq 0) = \mathbb{P}(X_1 > 0) = \frac{1}{2}$$ by symmetry. We also know:

$$\mathbb{P}(|X| \leq x) = \\ \mathbb{P}(-x \leq X \leq x) = \\ \mathbb{P}(X \leq x) - \mathbb{P}(X \leq -x) = \\ \mathbb{P}(X \leq x) - [1 - \mathbb{P}(X \leq x)] = \\ 2 \cdot \mathbb{P}(X \leq x) - 1$$

So, we try to say $$X_3 \sim \frac{1}{2} \cdot [2 \cdot \mathbb{P}(X \leq x) - 1] + \frac{1}{2} \cdot [-2 \cdot \mathbb{P}(X \leq x) + 1]$$, but the math here does not work out. Any help from this point would be appreciated.

R Code and Result

# Variables
N <- 1000000
X1 <- rnorm(n=N, mean=0, sd=1); X2 <- rnorm(n=N, mean=0, sd=1)
X3 <- rep(0, N)

# Compute X3
for (i in 1:N) {
if (X1[i] >= 0) {
X3[i] <- abs(X2[i])
} else {
X3[i] <- -1*abs(X2[i])
}
}

# Histogram (density)
hist(X3, freq=FALSE, breaks=50,
main="Histogram of X3 from Simulation",
xlab="Value of X3",
ylab="Density")
# Overlay standard normal for comparison
lines(density(X1), col="blue", lwd=2)


• $X_3$ has a mixture distribution: half the weight is given to a half-normal distribution and the other half of the weight is given to the negative of a half-normal distribution. Together these two halves describe a Normal distribution, QED. In general, the distribution of $X_3$ consists of the positive portion of the distribution of $X_2$, with weight $\Pr(X_1\ge 0)$, along with the reflected version of that positive portion, with the complementary weight. When $X_2$ is symmetric about $0$ and $X_1$ has probability $1/2$ of being non-negative, $X_3$ has the same distribution as $X_2.$
– whuber
Oct 23, 2018 at 2:43
• Crucial assumption of independence of $X_1$ and $X_2$ is missing in the question ( which is certainly mentioned in the original source of the problem). Oct 25, 2018 at 6:12

I assume $$X_1$$ and $$X_2$$ are independent of each other in what follows (thanks for pointing out the importance of this go to @stubbornatom.)

I would work directly with the cumulative density function of $$X_3$$:

$$P(X_3 \leq x) = 0.5[P(|X_2|\leq x) + P(-|X_2| \leq x)]$$

Consider $$x \leq 0$$. In this case, $$P(|X_2| \leq x) = 0$$, so, dropping it from the r.h.s. of the above equation, we have:

$$P(X_3 \leq x) = 0.5P(-|X_2| \leq x) = 0.5[P(X_2 \leq x) + P(-X_2 \leq x)]$$

where the second term on the r.h.s. covers the two cases $$X_2 \leq 0$$ (in which case $$-|X_2| = X_2$$) and $$X_2 > 0$$ (in which case $$-|X_2|=-X_2$$) respectively. Multiplying the terms of the last probability by $$-1$$ and switching the direction of the inequality results in:

$$P(X_3 \leq x) = 0.5[P(X_2 \leq x) + P(X_2 \geq -x)] = P(X_2 \leq x)$$

where the last equality is due to the symmetry of the standard Normal distribution around $$0$$.

A similar logic can be used to show that $$P(X_3 \leq x) = P(X_2 \leq x)$$ for the case $$x > 0$$. (Actually I think saying "symmetry" is probably sufficient.) Since the CDF of $$X_3$$ is everywhere equal to the CDF of $$X_2$$, and $$X_2$$ is standard Normal, it follows that $$X_3$$ is standard Normal too.

• This is exactly what I was looking for. Thank you for the detailed, step-by-step solution.
– ERT
Oct 23, 2018 at 1:14
• Didn't we assume that $X_1$ is independent of $X_2$? Oct 25, 2018 at 6:18
• @StubbornAtom - yes, that's why we can just substitute $0.5$ for $P(X_1 < 0)$ everywhere. Oct 25, 2018 at 14:13
• Was it apparent from the question? That's what I am asking. Oct 25, 2018 at 14:34
• No, it wasn't (+1), but without it you'd be hard-pressed to solve the problem, or so it seems to me. I'll add your point to the top of the anser, because it is important. Oct 25, 2018 at 16:09